Your proof is correct and it works for uncountable sets but I think that the second equality is a bit fast.
Since by definition of infinite sums, $\sum_{x \in E} 1=\sup \{\sum_{x \in F} 1:\, F\subset E,\, |F|<\infty\}=\text{cardinality} E$, if $\phi=\chi_{E}$, then $$\int_X\phi\,d\mu=\mu(E)=\sum_{x \in E} 1=\sum_{x \in X} \phi(x).$$
If $\phi$ is simple and nonnegative $\phi=\sum_{i}^nc_i\chi_{E_i}$, then $\int_X\phi\,d\mu=\sum_{{c_i>0}}^nc_i\mu(E_i)$ since $c_i\mu(E_i):=0$ if $c_i=0$. If one of the $E_i$ has infinitely many elements with $c_i>0$ then the integral is infinite and so is the infinite sum by the previous case. Otherwise each set $E_i$ with $c_i>0$ has finitely many elements and so by the previous step and the commutativity of finite sums $$\sum_{{c_i>0}}^nc_i\mu(E_i)=\sum_{{c_i>0}}^nc_i\sum_{x \in {X}}\chi_{E_i}(x)=\sum_{x \in {X}}\sum_{{c_i>0}}^nc_i\chi_{E_i}=\sum_{x \in {X}}\phi(x).$$
In the general case,
\begin{align} \int_X f \, d\mu & = \sup \{ \int_X \phi \, d\mu \ ,: \, 0 \le \phi \le f \text{ and $\phi$ simple } \} \\
&= \sup\Big\{\sum_{x \in {X}}\phi(x) \, :\, 0 \le \phi \le f \text{ and $\phi$ simple } \Big\} \\&= \sum_{x \in X} f(x),
\end{align}
where the last equality follows from the fact that $\sum_{x \in {X}}\phi(x)\le \sum_{x \in {X}}f(x)$ for each $\phi\le f$ and for every finite set $F$ we can take as simple function $\phi$ the function $f(x)$ if $x\in F$ and $0$ otherwise.
